A cube of side 10cm is filled with ice of density 0.9gm/cc. The thickness of the walls of the cube is 1 mm and thermal conductivity of the material of the cube is 0.01 cal/cm− o C. If the cube is placed in steam bath maintained at a temperature of 100 C , the time in which ice completely melts, is (latent heat of fusion of ice = 80 cal/gm.)
Answers
Answer:
Given data:
Length of the cube =10cm
Density of ice=0.9g/cc
Thickness of the walls of the cube =1mm
Thermal conductivity of the material of cube (K)=0.01cal/cm−
latent heat of fusion of ice =80cal/gm.
1Kcal=4184Joules
Now,
The mass of the ice in the cube is m=ρ×V=0.9×10
3
=900gm
So total heat absorbed is m×80cal/g=72×10
3
cal
As there are 6 walls , rate of heat conducted would be
Q = Q/t = K×Area×ΔT / L
⇒ Q =(6×0.01×100×100)/0.1=6000cal/s
So, time taken to melt the ice is Q / Qt =(72×10^3 )/6000=12sec
ans final answer will be = 12seconds.
thank you !
Explanation:
Given data:
Length of the cube =10cm
Density of ice=0.9g/cc
Thickness of the walls of the cube =1mm
Thermal conductivity of the material of cube (K)=0.01cal/cm−
o
C
latent heat of fusion of ice =80cal/gm.
1Kcal=4184Joules
Now,
The mass of the ice in the cube is m=ρ×V=0.9×10
3
=900gm
So total heat absorbed is m×80cal/g=72×10
3
cal
As there are 6 walls , rate of heat conducted would be
Q
.
=
t
Q
=
L
K×Area×ΔT
⇒ Q
.
=(6×0.01×100×100)/0.1=6000cal/s
So, time taken to melt the ice is
Q/t
Q
=(72×10
3
)/6000=12sec